Question
ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. Prove that P, Q, C and D are concyclic.
✓
Answer
ABCD is a parallelogram. A circle through A, B is so drawn that intersects AD at P and BC at Q. We have to prove that P, Q, C and D are concyclic. Join PQ.
Now, side AP of the cyclic quadrilateral APQB produced to D.
$\therefore\ \text{Ext. }\angle1=\text{int.opp. }\angle\text{B}$
$\because$ BA || CD and BC cuts them
$\therefore\angle\text{B}+\angle\text{C}=180^\circ$
[ $\because$ sum of int. $\angle\text{s}$ on the same side of the transversal is 180°]
or $\angle1+\angle\text{C}=180^\circ\ [\therefore\angle1=\angle\text{B}\ (\text{proved})]$
$\therefore$ PDCQ is cyclic quadrilateral.
Hence, the point P, Q, and are concyclic.
Now, side AP of the cyclic quadrilateral APQB produced to D.
$\therefore\ \text{Ext. }\angle1=\text{int.opp. }\angle\text{B}$
$\because$ BA || CD and BC cuts them
$\therefore\angle\text{B}+\angle\text{C}=180^\circ$
[ $\because$ sum of int. $\angle\text{s}$ on the same side of the transversal is 180°]
or $\angle1+\angle\text{C}=180^\circ\ [\therefore\angle1=\angle\text{B}\ (\text{proved})]$
$\therefore$ PDCQ is cyclic quadrilateral.
Hence, the point P, Q, and are concyclic.
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